C++(有点)工厂

Question

我看过很多关于 C++ 工厂的帖子，但到目前为止我还没有看到解决我的问题的解决方案。 (虽然我可能遗漏了一些东西。)

示例控制台应用程序:

#include <memory>
#include <map>
#include <iostream>

using namespace std;

class ResourceManager;

/// abstract base class
class Identity
{
public:
    int Id() const { return _id; }

    /// make this an abstract class
    virtual ~Identity() = 0 {}

protected:
    Identity() { _id = _nextId++; }

private:
    int _id;
    static int _nextId;
};

int Identity::_nextId = int();

/// derived classes
class Component : public Identity
{
    friend class ResourceManager;
public:
    ~Component() { }
};

class Entity : public Identity
{
    friend class ResourceManager;
public:
    ~Entity() { }
};


class ResourceManager
{
public:
    template<typename T>
    T& Create()
    {
        auto ptr = std::make_shared<T>();
        auto id = ptr->Id();

        _resources[id] = std::move(ptr);
        return *dynamic_pointer_cast<T>(_resources[id]);
    }

private:
    std::map<int, std::shared_ptr<Identity>> _resources;
};



int main(int argc, char *argv[])
{
    cout << "Factory test" << endl;

    ResourceManager r;

    auto& e = r.Create<Entity>();
    cout << "e.id = " << e.Id() << endl;

    Entity e2;
    cout << "e2.id = " << e2.Id() << endl;

    Component c;
    cout << "c.id = " << c.Id() << endl;

    std::getchar();
}

我需要确保只有 ResourceManager 可以实例化 Entity、 Component 以及从它们派生的任何类。

我已经查看并将 ResourceManager 作为友元类添加到 Identity， 并将构造函数设为私有(private)或 protected ，但没有成功。 (这可能是一条死胡同，或者只是我这边的一个实现问题。)

有什么建议吗？

更新和编辑

用可编译示例替换了代码。虽然 Identity() 的构造函数受到保护，但我仍然可以直接实例化派生类。

Answer 1

以下应该有效:

friend class ResourceManager; 应该在每个派生类中。( friend 不继承)。

class ResourceManager;

/// abstract base class
class Identity
{
public:
    int Id() const { return _id; }

    /// make this an abstract class
    virtual ~Identity() = 0;

    // Forbid any copy
    Identity(const Identity&) = delete;
    Identity(const Identity&&) = delete;
    Identity& operator = (const Identity&) = delete;
    Identity& operator = (Identity&&) = delete;

protected:
    Identity() { _id = _nextId++; }

private:
    int _id;
    static int _nextId;
};

// empty destructor
Identity::~Identity() {}

int Identity::_nextId = 0;

/// derived classes
class Component : public Identity
{
    friend class ResourceManager;
public:
    ~Component() { }
protected:
    Component() = default;
};

class Entity : public Identity
{
    friend class ResourceManager;
public:
    ~Entity() { }
protected:
    Entity() = default;
};


class ResourceManager
{
public:
    template<typename T>
    T& Create()
    {
         std::unique_ptr<T> ptr(new T);
         T& res = *ptr;

        _resources[ptr->Id()] = std::move(ptr);
        return res;
    }

    /// TODO: need to make sure that resource ID is actually of type T
    /// and that _resources contains ID.

    template<typename T>
    T* Get(int id)
    {
        auto it = _resources.find(id);
        if (it == _resources.end()) {
            return nullptr;
        }
        return dynamic_cast<T*>(it->second.get());
    }

private:
    std::map<int, std::unique_ptr<Identity>> _resources;
};

请注意，由于 ResourceManager 拥有我将 std::shared_ptr 更改为 std::unique_ptr 的资源。我用无效 ID 修复了 ResourceManager::Get。